Block sliding down frictionless semicircle

1. The problem statement, all variables and given/known data
A block of mass, m, sits atop a semicircular bowl of radius, r, and the angle the radius makes with the horizontal is [tex]\theta[/tex]. Find what angle, [tex]\theta[/tex], the block will slide off the bowl.

3. The attempt at a solution
Drawing a free-body diagram, the forces I think at work are the normal force, gravity, and possibly a centripetal force. The resolution of gravity along a circular path is where my trouble begins. I don't think it's really equivalent to resolve gravity as the sum of two x and y vectors along an incline, since this is a circle. Maybe resolving into x=rcos[tex]\theta[/tex] and y=rsin[tex]\theta[/tex] and saying mgrcos[tex]\theta[/tex]=X motion and Y motion is mgrsin[tex]\theta[/tex]=y. Then I'm stuck how to incorporate the normal force into either of those two equations (assuming they're accurate representations of the motion of this object along this semicircle). Maybe [tex]\sum{F_{y}}=F_{n}-mgrsin\theta=mar[/tex] and [tex]\sum{F_{x}}=mgrcos\theta=mar[/tex]. I guess if I could by some mathematical bastardization equate mgrcos[tex]\theta[/tex]=mgrsin[tex]\theta[/tex] I'd get that [tex]\theta=\frac{\Pi}{4}[/tex]

I hope you all can give me a little guidance so that I can solve this problem. I know it's been posted before, but I searched, and I couldn't any help toward a solution.

Staff: Mentor

so if the normal force equals zero then [tex]-mgrsin\theta=mgrcos\theta[/tex] and [tex]\theta=\frac{-\pi}{4}[/tex] which would have to become positive [tex]\frac{\pi}{4}[/tex] since...?

The normal force is zero as it loses contact with the surface, but why in the world are you attempting to equate "vertical" and "horizontal" components of gravity? (At least I think that's what you're doing. To properly find the components of gravity, draw yourself a diagram and apply a bit of trig. Don't convert coordinates using x = r cosθ.)